Question

A 1410-kg car moving east at 17.0 m/s collides with a 1880-kg car moving south at 15.0 m/s, and the two cars connect together.

a) What is the magnitude of the velocity of the cars right after the collision? (m/s)

b) What is the direction of the cars right after the collision? Enter the angle in degrees where positive indicates north of east and negative indicates south of east. (°)

c) How much kinetic energy was converted to another form during the collision? ( kJ)

Answer 1

Solve this problem using unit vector notations.

m1 = 1410 kg

u1 = 17.0 m/s (in east)

m2 = 1880 kg

u2 = 15.0 m/s (in south)

So, initial momentum of the two vehicles -

Pi = m1*u1 + m2*u2

= [(1410*17.0)i + (1880*15.0)(-j)] kg*m/s

= [23970i - 28200j] kg*m/s

After collision, the two cars stick togather.

Suppose, v is the velocity of the two cars after collision.

So, final momentum, Pf = (m1 + m2)*v

Apply conservation of momentum -

Pi = Pf

=> [23970i - 28200j] = (m1 + m2)*v

=> [23970i - 28200j] = (1410 + 1880)*v = 3290*v

=> v = sqrt[(23970^2 + (-28200)^2] / 3290 = 11.2 m/s (Answer)

(b) Direction of the final velocity, = tan^-1[(-28200) / 23970] = -49.6 deg

means, 49.6 deg. south of east.

(c) Total initial kinetic energy = (1/2)*m1*u1^2 + (1/2)*m2*u2^2

= 0.5*1410*17^2 + 0.5*1880*15^2

= 203745 + 211500 = 415245 J

Total final kinetic energy = (1/2)*(m1+m2)*v^2

= 0.5*3290*11.2^2 = 206348.8 J

Therefore, kinetic energy converted to another form = Total initial kinetic energy - Total final kinetic energy

= 415245 J - 206348.8 J = 208896.2 J (Answer)